Question

If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.14 m solution of this acid?

Answer 1

pH = 2.1

Let $HA (aq)$ resembles the acid; as a weak acid (a small value of $K_{a}$) $HA$ would partially dissociate to produce protons $H^{+}$ and $A^{-}$, its conjugate base. Let the final proton concentration (i.e., $[H^{+}]$) be $x$. (Apparently $x \ge 0$) Construct the following RICE table:

$\left\begin{array}{cccccc}\text{R}&HA(aq)&\rightleftharpoons&H^{+}(aq) &+ &A^{-}(aq)\\\text{I} & 0.14 \; \text{M} & &\\\text{C}&-x \; \text{M}& &+x \; \text{M} & & +x \; \text{M}\\E & (0.14 - x)\; \text{M} & & x \; \text{M} & &\+x \; \text{M}\end{array}\right$

By definition, (all concentrations are under equilibrium condition)

$\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right$

It is given that

$K_a = 5.4 \cdot 10^{-6}$

Equating and simplifying the two expressions gives a quadratic equation; solve the equation for $x$ gives:

$x^2 = 5.4 \cdot 10^{-6} \cdot (14 - x) \\x^2 + 5.4 \cdot 10^{-6} \cdot 14 \cdot x - 5.4 \cdot 10^{-6} \cdot 14 = 0 \\x = 0.0087 \; \text{M} \; (x \ge 0)$

The pH of a solutions equals the opposite of the logarithm of its proton concentration to base 10; thus for this particular solution

$\text{pH} = -\text{ln(}[H^{+}]\text{)} / \text{ln(}10\text{)} = 2.1$